Evaluating the contour integral $\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\mathrm dx$

Your coutour will work perfectly, so I wonder why you're hesitating to proceed with your calculation. Anyway, here is a solution:

Since $\sin^3 z = (\sin 3z - 3\sin z)/4$, we have

$$\int_{0}^{\infty} \frac{\sin^3 x}{x^3} \, dx = \frac{1}{8} \Im \lim_{\epsilon \to 0} \int_{\mathbb{R} \backslash (-\epsilon, \epsilon)} \frac{3 e^{iz} - e^{3iz} -2}{z^3} \, dz,$$

where the term $-2$ is introduced in order to cancel out the pole of order 3, without affecting the value of the integral. Consider the counterclockwise-oriented upper semicircle $C$ of radius $R$, centered at the origin, with semicircular indent of radius $\epsilon$. Let $\Gamma_{R}^{+}$ and $\gamma_{\epsilon}^{-}$ denote semicircular arcs of $C$ of radius $R$ and $\epsilon$, respectively. If we put

$$f(z) = \frac{3 e^{iz} - e^{3iz} - 2}{z^3},$$

then we find that

• On $\Gamma_R^+$, we have $|f(z)| \leq 6R^{-3}$ and thus $$\int_{\Gamma_{R}^{+}} f(z) \, dz \to 0 \quad \text{as } R \to \infty.$$

• Notice that $$ f(z) = \frac{3}{z} + O(1) \quad \text{near } z = 0. $$ So by the direct computation, $$\int_{\gamma_{\epsilon}^{-}} f(z) \, dz = -\int_{0}^{\pi} f(\epsilon e^{i\theta}) i\epsilon e^{i\theta} \, d\theta = -\int_{0}^{\pi} (3i + O(\epsilon)) \, d\theta \to -3\pi i \quad \text{as } \epsilon \to 0.$$ (This is exactly the same as $-\pi i$ times the residue of $f$ at $z = 0$. The emergence of residue can be attributed to the fact that $f$ has only simple pole at $z = 0$.)

Since $f(z)$ has no pole on the region enclosed by $C$, we have $$\lim_{\epsilon \to 0} \int_{\mathbb{R} \backslash (-\epsilon, \epsilon)} \frac{3 e^{iz} - e^{3iz} - 2}{z^3} \, dz = 3\pi i.$$ This proves the desired identity.

$$ {\rm J}\left(\alpha\right) \equiv. \int_{-\infty}^{\infty}{\sin^{3}\left(\alpha x\right) \over x^{3}}\,{\rm d}x\,, \qquad\qquad {\rm J}\left(0\right) = 0\,,\quad {\rm J}\left(1\right) = ? $$

$$ {\rm J}'\left(\alpha\right) = {3 \over 2}\int_{-\infty}^{\infty} {\sin\left(\alpha x\right)\sin\left(2\alpha x\right) \over x^{2}}\,{\rm d}x = {3 \over 4}\int_{-\infty}^{\infty} {\cos\left(\alpha x\right) - \cos\left(3\alpha x\right) \over x^{2}}\,{\rm d}x $$

$$ {\rm J}''\left(\alpha\right) = {3 \over 4}\int_{-\infty}^{\infty} {-\sin\left(\alpha x\right) + 3\sin\left(3\alpha x\right) \over x}\,{\rm d}x = {3 \over 4}\,2\pi\,{\rm sgn}\left(\alpha\right) = {3 \over 2}\,\pi\,{\rm sgn}\left(\alpha\right) $$

$$ {\rm J}'\left(\alpha\right) = {3 \over 2}\,\pi\left\vert\alpha\right\vert\,, \quad {\rm J}\left(1\right) = \int_{-\infty}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {3 \over 2}\,\pi\int_{0}^{1}\left\vert\alpha\right\vert\,{\rm d}\alpha = {3 \over 4}\,\pi $$

$$ \int_{0}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {1 \over 2}\,\int_{-\infty}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {3 \over 8}\,\pi $$